If the equation $\sqrt{a + \sqrt{x+a}} = x$ has exactly 2 solutions, then find the range of values of a

Question

I tried solving for a by squaring both sides but the equation turns out to be biquadratic which is out of my syllabus right now. What I think is about the use of inverse function and quadratic equations, but I am unable to apply these to solve. The answer to the problem is a ∈ $(-1/4,0)$

The below cited answer gives some hint to what I want to do

Aqua (https://math.stackexchange.com/users/463553/aqua), prove that for $-1/4 < a < 0$, the equation $\sqrt{a + \sqrt{x+a}} = x$ has exactly 2 solutions one of which is $(1+\sqrt{4a+1})/2$., URL (version: 2018-08-24): https://math.stackexchange.com/q/2893448

Answer 1

Fig. 1.: Curves of $f_a$ in red, $f_a^{-1}$ in blue, symmetrical with respect to bissecting line $y=x$ with from left to right $0,1,2,1$ common points. These curves are half-parabolas. This graphical representation shows how narrow is the range of values of $a$ for which there are 2 common points to these curves.

Let $$f_a(x)=y=\sqrt{x+a}\tag{1}$$

whose inverse function is $y=f_a^{-1}(x)=x^2-a\tag{2}$

(see figure 1).

Your issue

$$\text{There exists exactly 2 values of x such that } \ \sqrt{\sqrt{x+a}+a}=x$$

can be written as

$$\text{There exists exactly 2 values of x such that } \ f_a(\underbrace{f_a(x)}_{y \ge 0})=x\tag{3}$$

Besides,

$$f_a^{-1}(f_a(x))=x\tag{4}$$

Comparison between (3) and (4) amounts to say that:

$$\text{There exists exactly 2 (positive) values of y such that } \ f_a^{-1}(y)=f_a(y)\tag{5}$$

It is not our interest to work with equation (5) which gives a fourth degree equation.

An essential remark is that (5) is possible if and only if

$$ f_a^{-1}(y)=f_a(y) \color{red}{=y}\tag{6}$$

Indeed, when the curves associated to $f$ and $f^{-1}$ share a common point, this point belongs as well to the line bissector with equation $y=x$ (see figure 1).

(6) gives rise to a quadratic equation (only!)

$$x^2-a=x \ \iff \ x^2-x-a=0 \tag{7}$$

(we have switched back to variable $x$).

Condition "having 2 positive roots" for equation (7) is equivalent to the fact that the discriminant of (7) is positive and that moreover its smallest root is positive:

$$\Delta=1+4a>0 \ \text{and} \ \tfrac12(1-\sqrt{1+4a}) \ge 0 \ \iff \ \color{red}{a \in (-\tfrac14,0)}\tag{8}$$

(Check on fig. 1: the third case $a=-0.1$ is the only one with 2 intersection points).

Answer 2

Hint

Call $\sqrt{x+a}=y$ and then you have the system:

$$\sqrt{x+a}=y,\quad \sqrt{y+a}=x$$

where $x,y,x+a,y+a\ge0$.

Now square both equations and get $$x+a=y^2,\quad y+a=x^2.$$

Now subtract both equations and get $$x-y=y^2-x^2\Leftrightarrow (y-x)(y+x+1)=0.$$

So, you get two situations to analyse, $y=x$ or $x+y=-1$. Are both situations possible?

Can you finish?